I have this exercise:

Show that if $|a| < r <|b|$, then $\int_\gamma \! \frac{1}{(z-a)(z-b)} \, \mathrm{d}z=\frac{2\pi i}{a-b}$, where $\gamma$ denotes the circle centered at the origin, of radius $r$, with positive orientation.

I will use the parameterisation $re^{it}$ and integrate from $0$ to $2\pi$. But I get stuck. One way I tried was partial fractions, I know that :

$$\frac{1}{(z-a)(z-b)}=\frac{-1/(b-a)}{z-a}+\frac{1/(b-a)}{z-b},$$ but when I try to integrate the first part I get:

$$\int_0^{2\pi} \! \frac{-1/(b-a)}{re^{it}-a}rie^{it} \, \mathrm{d}t,$$ but how do I evaluate this? My first instinct is to use substitution, but it is not clear if I can when we are dealing with complex numbers. Any tips?

  • $\begingroup$ Do you know about Cauchy's integral formula? $\endgroup$ – Git Gud Sep 3 '14 at 16:07
  • $\begingroup$ @GitGud No sorry I do not I suspect I will learn it later. But this exercise is from a chapter where we haven't learned it yet, so I assume there is a way to evaulate the integral. $\endgroup$ – user119615 Sep 3 '14 at 16:08
  • $\begingroup$ Do you have any result about evaluating closed line integrals of functions whose singularities lie outside the closed curve? $\endgroup$ – Git Gud Sep 3 '14 at 16:11
  • $\begingroup$ @GitGud No sorry, the only result is that if the function has a primitive in an open region containing the curve, then the integral is 0. But I think the point here is that we do nto have a primitive. $\endgroup$ – user119615 Sep 3 '14 at 16:12
  • $\begingroup$ Mimic the last bullet of this answer after "A third alternative is to break up" . It's an elementary approach. What I do is break up the line in its upper and lower part. In doing this you can find an antiderivative for each part, etc. $\endgroup$ – Git Gud Sep 3 '14 at 16:13

Your partial fraction decomposition is the key: Note that $\frac{1}{z-b}$ does not have a pole inside $\gamma$, and hence its integral around $\gamma$ is $0$.

Now, for the integral of $\frac{1}{z-a}$ try the Cauchy integral formula.

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